Codeforces Round #334 B. More Cowbell （二分 + 贪心）

题意:

k个箱子装n个物品(n≤2∗k≤105),每个箱子最多装2个物品,求最小的满足要求的箱子的体积

分析:

小的能装大的也可以装,满足单调性可以二分,check的时候贪心一下,肯定是一大一小装才最好嘛,注意边界

代码:

////  Created by TaoSama on 2015-12-01//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, k;int a[N];bool check(int x) {    int l = 1, r = n, cnt = 0;    while(l <= r) {        if(a[r] > x) return false;        if(l != r) {            if(a[l] + a[r] > x) --r;            else ++l, --r;        } else ++l;        ++cnt;    }    return cnt <= k;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d", &n, &k) == 2) {        for(int i = 1; i <= n; ++i) scanf("%d", a + i);        int l = 1, r = 1e8;        while(l <= r) {            int m = l + r >> 1;            if(check(m)) r = m - 1;            else l = m + 1;        }        printf("%d\n", l);    }    return 0;}